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    (Original post by Brad0440)
    If you didn't divide by (-ΔS), then the '-' (negative) would stay with the T, making the equation: (G-ΔH)/ΔS=-T. Another general example for this would be if the equation was C=D-AB, then to make the equation equal to AB you would move the D over (C-D=-AB) and then divide by (-A) to make it equal B: (C-D)/-A=B. Again, if the negative wasn't moved over, then the equation would be (C-D)/A=-B.

    When you move the A in A+B=C, the sign of the A changes, because that is what is being moved over, so you get B=C-A. It would be like if you had 3+1=4, if you move the 1 over, you get 3=4-1, which is true. If it was the other way around (like B=A-C in the example), then the equation would become 3=1-4, which isn't true.

    Basically, if you are moving something over that is by itself (like A, B or ΔH), then the sign changes, but if you are dividing (or multiplying) by something to split up a term (like AB, or CD, or TΔS), then the sign stays the same.
    really, thank you so much ! I was having a mini panick attack before this....:clap2:
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    What conversions do we need to definitely know?


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    (Original post by AyshaK)
    It's kw = [oh-][h+] isnt it?

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    No I mean like if a reaction has H2O in e.g. H2O2 ==> H2O + 1/2O2 then do you include H2O in Kc? Cause there's one expression where apparently H2O is in so much excess it is considered to be constant and you don't include it

    =S
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    so wait, do we need to know the equation for dichromate, or just the manganese and the thiosulfate one?
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    (Original post by Sarangtaec)
    No I mean like if a reaction has H2O in e.g. H2O2 ==> H2O + 1/2O2 then do you include H2O in Kc? Cause there's one expression where apparently H2O is in so much excess it is considered to be constant and you don't include it

    =S
    you do include it in Kc, but it becomes Kw as conc of H20 is constant. The expression you are talking about is the Kstab value where H2O conc is large and constant and hence not needed.
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    Guys can someone explain to me how a buffer works??
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    I was wondering if anyone could help me with 6d in the January 2011 paper.

    'When dissolved in water, the enthalpy change of solution of the salt potassium fluoride, KF, is -15kJ mol-1.

    The salt rubidium fluoride, RbF, has an enthalpy change of solution in water of -24kJ mol-1.

    Suggest reasons for the difference between the enthalpy changes of solution of KF and RbF.'

    The thing that I'm confused about isn't directly asked about in the question, but it is still bugging me and I've spent ages trying to work it out. Why is the enthalpy change of solution of RbF more exothermic than that of KF? K+ has a smaller ionic radius than Rb+, so KF should have more exothermic/negative enthalpy change of hydration and lattice enthalpy values than RbF. I know the mark scheme says that lattice enthalpy affects delta H solution more than delta H hydration does, but that shouldn't make a difference, should it, as KF is more exothermic in both cases?
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    (Original post by Satta101)
    Can anyone help me with this please??!
    what do the answers say for that, because I'm not getting whole numbers, and I can't see where I've gone wrong...
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    (Original post by wl1)
    I was wondering if anyone could help me with 6d in the January 2011 paper.

    'When dissolved in water, the enthalpy change of solution of the salt potassium fluoride, KF, is -15kJ mol-1.

    The salt rubidium fluoride, RbF, has an enthalpy change of solution in water of -24kJ mol-1.

    Suggest reasons for the difference between the enthalpy changes of solution of KF and RbF.'

    The thing that I'm confused about isn't directly asked about in the question, but it is still bugging me and I've spent ages trying to work it out. Why is the enthalpy change of solution of RbF more exothermic than that of KF? K+ has a smaller ionic radius than Rb+, so KF should have more exothermic/negative enthalpy change of hydration and lattice enthalpy values than RbF. I know the mark scheme says that lattice enthalpy affects delta H solution more than delta H hydration does, but that shouldn't make a difference, should it, as KF is more exothermic in both cases?
    no they work against each other to produce enthalpy change of solution, sorry i can't explain it now - when i do the paper i'll tell you ( unless someone else explain it to you before! )
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    (Original post by tom2013)
    Guys can someone explain to me how a buffer works??
    a buffer is made up of a weak acid and its conjugate base - which could be obtained from the salt of the weak acid
    the weak acid partially dissociates
    the salt fully dissociates
    therefore, you have a high conc of weak acid and conjugate base, and few H+
    a buffer system is formed - which is basically the dissociation of the weak acid
    HA >< (my reversible arrows!) H+ + A-
    on addition of small amounts of alkali, the few H+ present will react to form H20 as H+ + OH- goes to H20. this will cause HA to dissociate, shifting the equilibrium to the right to replace the H+ ions that reacted
    on addition of small amounts of acid, this will react with the A- ions and so shift the equilibrium to the left, to decrease the excess H+ ions
    hope that helps!!
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    (Original post by wl1)
    I was wondering if anyone could help me with 6d in the January 2011 paper.

    'When dissolved in water, the enthalpy change of solution of the salt potassium fluoride, KF, is -15kJ mol-1.

    The salt rubidium fluoride, RbF, has an enthalpy change of solution in water of -24kJ mol-1.

    Suggest reasons for the difference between the enthalpy changes of solution of KF and RbF.'

    The thing that I'm confused about isn't directly asked about in the question, but it is still bugging me and I've spent ages trying to work it out. Why is the enthalpy change of solution of RbF more exothermic than that of KF? K+ has a smaller ionic radius than Rb+, so KF should have more exothermic/negative enthalpy change of hydration and lattice enthalpy values than RbF. I know the mark scheme says that lattice enthalpy affects delta H solution more than delta H hydration does, but that shouldn't make a difference, should it, as KF is more exothermic in both cases?
    i also didn't get this for the same reasons - anyone?!
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    (Original post by georgiaaaxo)
    a buffer is made up of a weak acid and its conjugate base - which could be obtained from the salt of the weak acid
    the weak acid partially dissociates
    the salt fully dissociates
    therefore, you have a high conc of weak acid and conjugate base, and few H+
    a buffer system is formed - which is basically the dissociation of the weak acid
    HA >< (my reversible arrows!) H+ + A-
    on addition of small amounts of alkali, the few H+ present will react to form H20 as H+ + OH- goes to H20. this will cause HA to dissociate, shifting the equilibrium to the right to replace the H+ ions that reacted
    on addition of small amounts of acid, this will react with the A- ions and so shift the equilibrium to the left, to decrease the excess H+ ions
    hope that helps!!


    thank you soo much, is that all you need to know?? Soo I just apply that to the questions??
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    (Original post by MedMed12)
    [/B]

    Nooooooooooo!! Anything but Jan
    JAN 2013 was horrible. Hope it's nothing like that
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    (Original post by otrivine)
    _
    Worked example:
    Attached Images
     
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    (Original post by otrivine)
    thats what I got, the mark scheme made a mistake
    Oh ok, I don't even have the mark scheme just what the teacher corrected but I worked out how to get the final answer now.

    Also for the specimen paper, the Kc question. It's 3b (ii) I keep getting 7.28 but answer is supposed to be 14.6 why is that?
    I divided the no of moles at eqm by 2.0dm^3 as that's its volume. The only way I see of getting to 14.6 is to multiply by 2 my answer of 7.28.
    But is that correct & why would I have to do so?
    Any ideas?
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    (Original post by wl1)
    I was wondering if anyone could help me with 6d in the January 2011 paper.

    'When dissolved in water, the enthalpy change of solution of the salt potassium fluoride, KF, is -15kJ mol-1.

    The salt rubidium fluoride, RbF, has an enthalpy change of solution in water of -24kJ mol-1.

    Suggest reasons for the difference between the enthalpy changes of solution of KF and RbF.'

    The thing that I'm confused about isn't directly asked about in the question, but it is still bugging me and I've spent ages trying to work it out. Why is the enthalpy change of solution of RbF more exothermic than that of KF? K+ has a smaller ionic radius than Rb+, so KF should have more exothermic/negative enthalpy change of hydration and lattice enthalpy values than RbF. I know the mark scheme says that lattice enthalpy affects delta H solution more than delta H hydration does, but that shouldn't make a difference, should it, as KF is more exothermic in both cases?
    Enthalpy change of solution consists of the energy required to break down the ionic lattice into it's isolated ions (-lattice enthalpy), and the enthalpy of hydration. So although KF has the more exothermic hydration, the fact that KF's enthalpy of solution is less exothermic, must mean that the breakdown of the ionic lattice requires more energy - is more endothermic, and outweighs the more exothermic hydration.

    In that question they talk about it in terms of lattice enthalpy, you had to explain that K had a smaller ionic radius, so more negative lattice enthalpy. And then say lattice enthalpy must have a larger impact on enthalpy of solution.
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    (Original post by Brad0440)
    Worked example:
    THANK you, and if it was not feasible , it would have been Cr+ what
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    Can someone explain cis- platin please?
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    (Original post by tom2013)
    thank you soo much, is that all you need to know?? Soo I just apply that to the questions??
    i think so yeah! i don't remember any other theory about buffers we need to know at least..and then of course you need to know how to do the calculations for ph and the acid/conj base ratios
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    (Original post by otrivine)
    THANK you, and if it was not feasible , it would have been Cr+ what
    That reaction will always be feasible (under standard conditions). The question might ask something like is the reaction between Cr and Cu feasible under standard conditions? The answer would be no because Cr is a reactant and Cu is a product, so that reaction will not occur (it isn't feasible).
 
 
 
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